Gauss-Lucas Theorem

Theorem

Let $P$ be a (nonconstant) polynomial with complex coefficients.

Then all zeros of its derivative $P'$ belong to the convex hull of the set of zeros of $P$.

Proof

Over the complex numbers, $P$ is a product of prime factors:

$\ds \map P z = \alpha \prod_{i \mathop = 1}^n (z-a_i)$

where:

the complex numbers $a_1, a_2, \ldots, a_n$ are the (not necessary distinct) zeros of the polynomial $P$

the complex number $\alpha$ is the leading coefficient of $P$

$n$ is the degree of $P$.

Let $z$ be any complex number for which $\map P z \ne 0$.

Then we have for the logarithmic derivative:

$\ds \frac {\map {P'} z} {\map P z} = \sum_{i \mathop = 1}^n \frac 1 {z - a_i}$

In particular, if $z$ is a zero of $P'$ and $\map P z \ne 0$, then:

$\ds \sum_{i \mathop = 1}^n \frac 1 {z - a_i} = 0$

or:

$\ds \sum_{i \mathop = 1}^n \frac {\overline z - \overline {a_i} } {\size {z - a_i}^2} = 0$

This may also be written as:

$\ds \paren {\sum_{i \mathop = 1}^n \frac 1 {\size {z - a_i}^2} } \overline z = \paren {\sum_{i \mathop = 1}^n \frac 1 {\size {z - a_i}^2} \overline {a_i} }$

Taking their conjugates, we see that $z$ is a weighted sum with positive coefficients that sum to one, or the barycenter on affine coordinates, of the complex numbers $a_i$ (with different mass assigned on each root whose weights collectively sum to $1$).

If $\map P z = \map {P'} z = 0$, then:

$z = 1 \cdot z + 0 \cdot a_i$

and is still a convex combination of the roots of $P$.

$\blacksquare$

Source of Name

This entry was named for Carl Friedrich Gauss and François Édouard Anatole Lucas.